Question

Calculate the number of turns per unit length in a solenoid when the charge of $4\text{C}$, flows for $2\text{minutes}$, producing a magnetic field of $25\times {10}^{-5}\text{T}$.

• A. $3000$
• B. $6000$
• C. $8000$
• D. $12000$

Answer 1

The correct option is B $6000$

Given,
$q=4\text{C},t=2\text{min}=2\times 60=120\text{Sec}$
$B=25\times {10}^{-5}\text{T}$
$n=?$

$n=\frac{B}{{\mu }_{0}i}=\frac{25\times {10}^{-5}}{4\pi \times {10}^{-7}\times \left(\frac{4}{120}\right)}$

$\therefore n=\frac{25\times 120\times {10}^{-5}}{16\times 3.14\times {10}^{-7}}$

$\Rightarrow n=5971.3\approx 6000\text{turns per unit length}$

Why this Question ? Note: Current flowing per unit time is current. Also, the magnetic field due to a solenoid (ideal) is given by, $B={\mu }_{0}ni$